Question 23.1: For a binary liquid mixture of components 1 and 2, if the pa...
For a binary liquid mixture of components 1 and 2, if the partial molar enthalpy \overline{ H }_{1} is known as a function of T, P, and x _{1}, calculate \overline{ H }_{2} \text { and } H. The data \overline{ H }_{1}= f \left( x _{1}\right) is available at constant T and P.
We begin by writing the generalized Gibbs-Duhem relation for B = H at constant T and P, derived in Lecture 22, which for components 1 and 2, is given by:
x _{1} d \overline{ H }_{1}+ x _{2} d \overline{ H }_{2}=0 (23.1)
Because we are given \overline{ H }_{1}=\overline{ H }_{1}\left( T , P , x _{1}\right), we differentiate Eq. (23.1) with respect to x _{ 1 }, at constant T and P. This yields:
x _{1}\left(\frac{\partial \overline{ H }_{1}}{\partial x _{1}}\right)_{ T , P }+ x _{2}\left(\frac{\partial \overline{ H }_{2}}{\partial x _{1}}\right)_{ T , P }=0 (23.2)
where x_{2}=1-x_{1}. Rearranging Eq. (23.2) yields:
\left(\frac{\partial \overline{ H }_{2}}{\partial x _{1}}\right)_{ T , P }=-\frac{ x _{1}}{1- x _{1}}\left(\frac{\partial \overline{ H }_{1}}{\partial x _{1}}\right)_{ T , P } (23.3)
where the partial derivative of \overline{ H }_{1} with respect to x _{ 1 } , at constant T and P, is known as a function of x _{ 1 }.
Integrating Eq. (22.3) from x _{1}=0 (pure component 2) to x _{ 1 } yields:
\begin{aligned}\int_{0}^{ x _{1}}\left(\frac{\partial \overline{ H }_{2}}{\partial x _{1}}\right)_{ T , P } dx _{1} &=\int_{0}^{ x _{1}}\left( d \overline{ H }_{2}\right)_{ T , P }=\overline{ H }_{2}\left( T , P , x _{1}\right)-\overline{ H }_{2}( T , P , 0) \\=&-\int_{0}^{ x _{1}}\left(\frac{ x _{1}}{1- x _{1}}\right)\left(\frac{\partial \overline{ H }_{1}}{\partial x _{1}}\right)_{ T , P } dx _{1}\end{aligned} (23.4)
where \overline{ H }_{2}( T , P , 0)= H _{2}( T , P ) is the molar enthalpy of pure component 2. Rearranging Eq. (23.4) yields:
\overline{ H }_{2}\left( T , P , x _{1}\right)= H _{2}( T , P )-\int_{0}^{ x _{1}}\left(\frac{ x _{1}}{1- x _{1}}\right)\left(\frac{\partial \overline{ H }_{1}}{\partial x _{1}}\right)_{ T , P } dx _{1} (23.5)
Finally, given \overline{ H }_{1}\left( T , P , x _{1}\right), and having calculated \overline{ H }_{2}\left( T , P , x _{1}\right) using Eq. (23.5), we can “assemble” H as follows:
H = x _{1} \overline{ H }_{1}+ x _{2} \overline{ H }_{2} (23.6)
Equation (23.6) shows that if we know (n-1) \bar{B}_{i} s, we can compute the nth partial molar B and then assemble B. In this example, B = H, and n = 2.